Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$ Why does the following hold:
\begin{equation*}
\displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ?
\end{equation*}
Can we generalize the above to 

$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?

Are there some values of $x$ for which the above formula is invalid?
What about if we take only a finite number of terms? Is there a simpler formula?

$\displaystyle \sum_{n=0}^{N} x^n$

Is there a name for such a sequence?

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 A: If u expand your summation you get series $$1+0.7+0.7^2+\dots$$ as it is geometric series (you can see it here http://en.wikipedia.org/wiki/Geometric_series) $$\sum_{n=0}^{\infty}{0.7^n}=\frac{1}{1-0.7}$$
Or 
$S=1+x+x^2+\dots$
$xS_n=x+x^2+\dots=S-1$
now u take
$S-xS=1$
$S(1-x)=1
\implies S=\frac{1}{1-x}$
here your $x=0.7$
A: I find the  proof here lovely.
A: it's called a geometric series.  let $-1<x<1$ and let $S_n=1+x+x^2+...+x^n$.  then $xS_n=x+...+x^{n+1}=S_n-1+x^{n+1}$.  move stuff around to get $$S_n=\frac{1-x^{n+1}}{1-x}$$ take the limit as $n\to\infty$ (noting that $x^n\to0$ if $|x|<1$)
A: By binomial theorem,
$$\frac{1}{1-x}=(1-x)^{-1}\\=1+{-1\choose1}(-x)+{-1\choose2}(-x)^2+{-1\choose3}(-x)^3...\\=1+x+x^2+x^3+...,$$ 
which is a geometric series.
A: For all $x\neq1$, we have $$\sum_{k=0}^n x^k=1+x+x^2+...+x^n=\frac{(1+x+x^2+...+x^n)(1-x)}{1-x}=\frac{1-x^{n+1}}{1-x}$$
In the case $x=1$, the sum is $\underbrace{1+...+1}_{n+1\text{}}=n+1$.
Now if $|x|<1$, we have
$$\sum_{k=0}^\infty x^k=\lim_{n\uparrow \infty}\sum_{k=0}^n x^k=\lim_{n\uparrow \infty}\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}.$$
This infinite series is a geometric series and is convergent if and only if $|x|<1.$
A: Using the fact that $$x^n - y^n=(x-y)\sum_{k=1}^nx^{k-1}y^{n-k}$$
Set $x=1$ and $y \in \mathbb{N}$ to get the desired result.
$$1 - y^n=(1-y)\sum_{k=1}^ny^{n-k} \iff \sum_{k=1}^ny^{n-k}=y^n\sum_{k=1}^ny^{-k}=\frac{y^n-1}{y-1} \iff \sum_{k=1}^n \frac{1}{y^k} = \frac{y^n-1}{y^{n+1}-y^n}$$
Now, using the limit of the result we have $$\large{\lim_{n\to \infty}\frac{y^n-1}{y^{n+1}-y^n} = \frac{1}{y-1}}$$.
Some examples,
$$\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k = 1$$
$$\sum_{k=1}^\infty \left(\frac{1}{3}\right)^k = 1/2$$
$$\sum_{k=1}^\infty \left(\frac{1}{4}\right)^k = 1/3$$
$$\sum_{k=1}^\infty \left(\frac{1}{5}\right)^k = 1/4$$
Also, if $y \to \infty \implies $
$$\sum_{k=1}^\infty \left(\frac{1}{y}\right)^k = 0$$
